Derive the kinematic equations of motion for constant acceleration.

Explanation

(i) Consider an object moving in a straight line with uniform or constant acceleration 'a'.
(ii) Let 'u' be the initial velocity at time t=0 and 'v' be the final velocity at time t.
(1) Velocity - time relation:
(i) Acceleration, `a=(dv)/(dt)` or dv = a dt
(ii) By integrating both sides, we get,
`int_(u)^(v)dv=int_(0)^(t)adt=aint_(0)^(t)dt=a[t]_(0)^(t)`
`v-u=at`
`v=u+at`
(2) Displacement - time relation:
(i) Velocity, `v=(ds)/(dt)`
or `ds=vdt=(u+at)dt`
`" "[becausev=u+at]`
(ii) By integrating both sides, we get,
`int_(0)^(s)ds=int_(0)^(t)(u+at)dt`
`int_(0)^(s)ds=uint_(0)^(t)dt+aint_(0)^(t)tdt`
`s=ut+1/2at^(2)`
Velocity - displacement relation:
(i) Acceleration `a=(dv)/(dt)=(dv)/(ds)(ds)/(dt)=(dv)/(ds)v`
`ds=1/avdv`
(ii) By integrating both sides, we get,
`int_(0)^(s)ds=1/aint_(u)^(v)vdv=1/a[u^(2)/2]_(u)^(v)`
`v^(2)=u^(2)+2as`
Alternative method :
`a=vdv(dv)/(ds), ads=vdv`
Integrating both sides `int_(0)^(s)ads=int_(u)^(v)vdv`
If acceleration is constant, `aint_(0)^(s)ads=int_(u)^(v)vdv`
`a[s]_(0)^(s)={v^(2)/2}_(u)^(v)=((v^(2)-u^(2)))/2, 2as=v^(2)-u^(2)`
`v^(2)=u^(2)+2as`
(4) Displacement - average velocity relation :
(i) Final Velocity, `v=u+at`
`" "at=v-u " ...(1)"`
(ii) We know displacement `s=ut+1/2at^(2)`
(iii) Substituting equation (1), we get,
`s=ut+1/2(v-u)t, s=ut+1/2vt-1/2ut`
`s=((u+v)t)/2`